Question

If \[11{z^{10}} + 10i{z^9} + 10iz - 11 = 0\], then what will be the possible value of \[\left| z \right|\].

Answer 1

Hint: The given function is a polynomial of degree \[10\] and so it is analytic and the equation has \[10\] roots. One can solve this equation by rearranging the equation by keeping the higher order on one side. Then we can prove by showing the contradiction \[\left| z \right| < 1\]or \[\left| z \right| > 1\]. Finally we will get the required result.

Complete step-by-step solution:
Consider the given equation \[11{z^{10}} + 10i{z^9} + 10iz - 11 = 0\].
Rearranging the equation by keeping the higher order on one side and the remaining on the other side.
\[{z^9}\left( {11z + 10i} \right) = 11 - 10iz\]
Taking the term \[11z + 10i\] to other side,
\[ \Rightarrow {z^9} = \dfrac{{11 - 10iz}}{{11z + 10i}} - - - - \left( 1 \right)\]
Let \[z\] be a complex number that satisfies the equation \[11{z^{10}} + 10i{z^9} + 10iz - 11 = 0\], \[z = x + iy\], we can find the roots by considering two cases.
Case 1:
For \[\left| z \right| < 1\],
\[ \Rightarrow \left| {x + iy} \right| = \sqrt {\left( {x + iy} \right)\left( {x - iy} \right)} \]
\[ \Rightarrow \sqrt {{x^2} + {y^2}} < 1\]
Taking modulus on both sides of the equation (1), we get
\[\left| {{z^9}} \right| = \left| {\dfrac{{11 - 10iz}}{{11z + 10i}}{\text{ }}} \right|\]
Substitute the value of z,
\[\left| {{z^9}} \right| = \left| {\dfrac{{11 - 10i\left( {x + iy} \right)}}{{11\left( {x + iy} \right) + 10i}}} \right|\]
Expanding further we get,
\[\left| {{z^9}} \right| = \left| {\dfrac{{11 - 10ix + 10y}}{{11x + 11iy + 10i}}} \right| - - - - \left( 2 \right)\]
We know that \[\left| {a + ib} \right| = \sqrt {\left( {a + ib} \right)\left( {a - ib} \right)} = \sqrt {{a^2} + {b^2}} \]
Consider the numerator \[\left| {11 - 10ix + 10y} \right|\],
Let \[a = 11 + 10y,b = - 10x\]
\[\left| {11 - 10ix + 10y} \right| = \sqrt {{{\left( {10y + 11} \right)}^2} + {{\left( { - 10x} \right)}^2}} \]
\[ \Rightarrow \sqrt {{{\left( {10y + 11} \right)}^2} + 100{x^2}} - - - - \left( 3 \right)\]
Consider the denominator \[\left| {11x + 11iy + 10i} \right|\],
Let \[a = 11x,b = 11y + 10\]
\[\left| {11x + 11iy + 10i} \right| = \sqrt {{{\left( {11x} \right)}^2} + {{\left( {11y + 10} \right)}^2}} \]
\[ \Rightarrow \sqrt {{{\left( {11y + 10} \right)}^2} + 121{x^2}} - - - - \left( 4 \right)\]
Substituting equation (3) and (4) in (2),
\[\left| {{z^9}} \right| = \dfrac{{\sqrt {{{\left( {10y + 11} \right)}^2} + 100{x^2}} {\text{ }}}}{{\sqrt {{{\left( {11y + 10} \right)}^2} + 121{x^2}} }}\]
Expanding sum of squares on both numerator and denominator, we get
\[ \Rightarrow \left| {{z^9}} \right| = \dfrac{{\sqrt {100{y^2} + 220y + 121 + 100{x^2}} {\text{ }}}}{{\sqrt {121{y^2} + 220y + 100 + 121{x^2}} }}\]
Hence,
\[ \Rightarrow \left| {{z^9}} \right| = \dfrac{{\sqrt {100\left( {{x^2} + {y^2}} \right) + 220y + 121} {\text{ }}}}{{\sqrt {121\left( {{x^2} + {y^2}} \right) + 220y + 100} }}\]
\[\because {x^2} + {y^2} < 1\]
Rearranging the terms we get,
\[\therefore 100\left( {{x^2} + {y^2}} \right) + 220y + 121 > 121\left( {{x^2} + {y^2}} \right) + 220y + 100\]
Hence,
\[ \Rightarrow \left| {{z^9}} \right| > 1\]
\[ \Rightarrow \left| z \right| > 1\]
This is a contradiction.
Case 2:
For \[\left| z \right| > 1\],
\[ \Rightarrow \left| {x + iy} \right| = \sqrt {\left( {x + iy} \right)\left( {x - iy} \right)} \]
\[ \Rightarrow \sqrt {{x^2} + {y^2}} > 1\]
Taking modulus on both sides of the equation (1), we get
\[\left| {{z^9}} \right| = \left| {\dfrac{{11 - 10iz}}{{11z + 10i}}{\text{ }}} \right|\]
Substitute the value of z,
\[\left| {{z^9}} \right| = \left| {\dfrac{{11 - 10i\left( {x + iy} \right)}}{{11\left( {x + iy} \right) + 10i}}} \right|\]
Expanding further we get,
\[\left| {{z^9}} \right| = \left| {\dfrac{{11 - 10ix + 10y}}{{11x + 11iy + 10i}}} \right| - - - - \left( 5 \right)\]
We know that \[\left| {a + ib} \right| = \sqrt {\left( {a + ib} \right)\left( {a - ib} \right)} = \sqrt {{a^2} + {b^2}} \]
Consider the numerator \[\left| {11 - 10ix + 10y} \right|\],
Let \[a = 11 + 10y,b = - 10x\]
\[\left| {11 - 10ix + 10y} \right| = \sqrt {{{\left( {10y + 11} \right)}^2} + {{\left( { - 10x} \right)}^2}} \]
\[ \Rightarrow \sqrt {{{\left( {10y + 11} \right)}^2} + 100{x^2}} - - - - \left( 6 \right)\]
Consider the denominator \[\left| {11x + 11iy + 10i} \right|\],
Let \[a = 11x,b = 11y + 10\]
\[\left| {11x + 11iy + 10i} \right| = \sqrt {{{\left( {11x} \right)}^2} + {{\left( {11y + 10} \right)}^2}} \]
\[ \Rightarrow \sqrt {{{\left( {11y + 10} \right)}^2} + 121{x^2}} - - - - \left( 7 \right)\]
Substituting equation (6) and (7) in (5),
\[\left| {{z^9}} \right| = \dfrac{{\sqrt {{{\left( {10y + 11} \right)}^2} + 100{x^2}} {\text{ }}}}{{\sqrt {{{\left( {11y + 10} \right)}^2} + 121{x^2}} }}\]
Expanding sum of squares on both numerator and denominator, we get
\[ \Rightarrow \left| {{z^9}} \right| = \dfrac{{\sqrt {100{y^2} + 220y + 121 + 100{x^2}} {\text{ }}}}{{\sqrt {121{y^2} + 220y + 100 + 121{x^2}} }}\]
Simplifying we get,
\[ \Rightarrow \left| {{z^9}} \right| = \dfrac{{\sqrt {100\left( {{x^2} + {y^2}} \right) + 220y + 121} {\text{ }}}}{{\sqrt {121\left( {{x^2} + {y^2}} \right) + 220y + 100} }}\]
\[\because {x^2} + {y^2} > 1\]
Rearranging the terms we get,
\[\therefore 100\left( {{x^2} + {y^2}} \right) + 220y + 121 > 121\left( {{x^2} + {y^2}} \right) + 220y + 100\]
Hence,
\[ \Rightarrow \left| {{z^9}} \right| < 1\]
\[ \Rightarrow \left| z \right| < 1\]
This is a contradiction.
$\therefore $ \[\left| z \right| = 1\], which means all the roots lie on the circle.

Note: In logic and mathematics, proof by contradiction is a form of proof that establishes the truth or the validity of a proposition, by showing that assuming the proposition to be false leads to a contradiction. Proof by contradiction is also known as indirect proof, proof by assuming the opposite.